Notes on Cold Atoms

Here is an elementary introduction to Cold Atoms:
Part-One: A crash course on COLD ATOMS, BEC-BCS crossover (pdf)
Part-Two: Topology in cold atoms (topological superfluid, Haldane insulator), Into the Quantum Hall effect (pdf)

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Notes on Cold Atoms:

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1.Scattering Theory
1.1 Classical Hard-Sphere Scattering
Consider a particle scattered by a hard-sphere target with radius $a$. The impact factor is
\[
b(\theta)=\sin\left(\frac{\pi-\theta}{2}\right)a,
\]
the total cross section is
\begin{align*}
\sigma&=\int_0^a2\pi b(\theta)db(\theta)\\
&=-\int_\pi^0\pi a^2\cos(\theta/2)\sin(\theta/2)d\theta\\
&=-\int_\pi^0\pi a^2/2\sin(\theta)d\theta\\
&=\pi a^2,
\end{align*}
which is obvious from classical point of view.

 

2.Feshbach Resonance

2.1 The Nozi`eres and Schmitt-Rink (NSR) descriptio of the BCS-BEC crossover
We consider a system with two-component fermions label as $\vert\uparrow\rangle$ and $\vert\downarrow\rangle$ states, the Hamiltonian of the system is
\begin{eqnarray}
H=\sum_{k,\sigma=\uparrow, \downarrow}\epsilon_kc_{k\sigma}^\dagger c_{k\sigma}+g\sum_{kpq}c_{k/2+p\uparrow}^\dagger c_{k/2-p\downarrow}^\dagger c_{k/2-q\downarrow}c_{k/2+q\uparrow}.
\end{eqnarray}
Now we want to get a perturbative solution of the thermodynamic potential. This can be done as follows.

We first scale the interaction part with $\lambda$, and the corresponding partition function becomes
\begin{eqnarray}
Z_\lambda=\rm{Tr}\left[e^{-\beta\left(H_0+\lambda V\right)}\right].
\end{eqnarray}
Then we calculate the derivative of the thermodynamic potential
\begin{eqnarray}
\frac{\partial\Omega_\lambda}{\partial\lambda}=-\frac{1}{\beta}\frac{\partial\rm{ln}Z_\lambda}{\partial\lambda}=\frac{1}{\lambda}\langle\lambda V\rangle.
\end{eqnarray}
So we have the thermodynamic potential
\begin{eqnarray}
\Omega_\lambda-\Omega_0=\int_0^1\frac{1}{\lambda}\langle\lambda V\rangle d\lambda,
\end{eqnarray}
where $\Omega_0=-\rm{ln}Z_0/\beta=-\rm{ln}\rm{Tr}\left[e^{-\beta H_0}\right]$ is the thermodynamic potential for the non-interaction system.

We can carry out diagrammatic perturbation as in the figure with
\begin{eqnarray}
\langle\lambda V\rangle=-\frac{1}{\beta}\sum_{k, \omega}\chi(k,\omega) T(k,\omega),
\end{eqnarray}
where
\begin{eqnarray}
\chi(k,\omega_n=2\pi ni/\beta)&&=-\frac{1}{\beta}\sum_{q,\omega_m}\frac{1}{\omega_n-\omega_m-\epsilon_{k/2+q}}\frac{1}{\omega_m-\epsilon_{k/2-q}}\\
&&=\frac{1}{2\pi i}\sum_q\int\frac{1}{e^{i\beta\omega}+1}\frac{1}{\omega_n-\omega_m-\epsilon_{k/2+q}}\frac{1}{\omega_m-\epsilon_{k/2-q}}d\omega\\
&&=\sum_q\frac{1-f_{k/2+q}-f_{k/2-q}}{\omega_n-\epsilon_{k/2+q}-\epsilon_{k/2-q}},
\end{eqnarray}
with $\epsilon_k=k^2-\mu$ and $f_k=f_{\epsilon_k}$ is the Fermi distribution. The ladder approximation of the $T$-matrix gives
\begin{eqnarray}
T(k,\omega)=g+g\chi(k,\omega)T(k,\omega),
\end{eqnarray}
with the solution
\begin{eqnarray}
T(k,\omega)=\frac{g}{1-g\chi(k,\omega)}.
\end{eqnarray}
So we have the thermodynamic potential
\begin{eqnarray}
\Omega_\lambda-\Omega_0=\int_0^1\frac{1}{\lambda}\langle\lambda V\rangle d\lambda=-\frac{1}{\beta}\sum_{k,\omega}\frac{g\chi}{1-g\chi}d\lambda=\frac{1}{\beta}\sum_{k,\omega}{\rm ln} \left(1-g\chi\right).
\end{eqnarray}
Define the phase shift
\begin{eqnarray}
\delta(k,\omega)=-{\rm Im}\left[{\rm ln}\left(1-g\chi\right)\right],
\end{eqnarray}
and by choosing the integral contour we have
\begin{eqnarray}
\Omega-\Omega_0=\frac{1}{2\pi i}\sum_k\int\frac{1}{e^{i\beta\omega}-1}{\rm ln}\left(1-g\chi\right)d\omega=-\frac{1}{\pi}\sum_k\int_{-\infty}^{\infty}\frac{1}{e^{\beta\omega}-1}\delta(k,\omega)d\omega.
\end{eqnarray}

Now we can see the superfluid transition corresponds to the divergence of the $T$-matrix, i.e.
\begin{eqnarray}
1-g\chi(k=0,\omega=0)=0.
\end{eqnarray}
For a canonical system with conserved particle numbers, we have
\begin{eqnarray}
N-N_0=-\frac{\partial\left(\Omega-\Omega_0\right)}{\partial \mu}=\frac{1}{\pi}\sum_k\int_{-\infty}^\infty\frac{1}{e^{\beta\omega}-1}\frac{\partial\delta(k,\omega)}{\partial\mu}d\omega.
\end{eqnarray}

3.Spinor Gases

For a system with spin-independent two-body interaction $V$, the total spin $F$ of the two atoms is a good quantum number, and we can expand the interaction as
\begin{eqnarray}
V&&=\sum_{F}g_F\vert F\rangle\langle F\vert\\
&&=\sum_{F;m_1,m_2;m’_1,m’_2}g_F\vert m_1m_2\rangle\langle m_1m_2\vert F\rangle\langle F\vert m’_1m’_2\rangle\langle m’_1m’_2\vert\\
&&=\sum_{F;m_1,m_2;m’_1,m’_2}g_FC^F_{m_1,m_2;m’_1m’_2}\psi^\dagger_{m_1}\psi^\dagger_{m_2}\psi_{m’_2}\psi_{m’_1}.
\end{eqnarray}
For $s$-wave scattering, the orbital wavefunction of two scattering atoms is symmetric, thus for both bosonic and fermionic systems the total spin can only allow $F=0,2,4,$ etc. For example, we have the interaction of a spin-1 Bose gases as
\begin{eqnarray}
V=\frac{1}{2}&&\left[g_2\left(\psi^\dagger_{+1}\psi^\dagger_{+1}\psi_{+1}\psi_{+1}+\psi^\dagger_{-1}\psi^\dagger_{-1}\psi_{-1}\psi_{-1}+2\psi^\dagger_{+1}\psi^\dagger_{0}\psi_{+1}\psi_{0}+2\psi^\dagger_{-1}\psi^\dagger_{0}\psi_{-1}\psi_{0}\right)\right.\\
&&\left.+\frac{g_0+2g_2}{3}\psi^\dagger_{0}\psi^\dagger_{0}\psi_{0}\psi_{0}+\frac{2}{3}\left(2g_0+g_2\right)\psi^\dagger_{+1}\psi^\dagger_{-1}\psi_{+1}\psi_{-1}\right.\\
&&\left.+\frac{2}{3}\left(g_2-g_0\right)\left(\psi^\dagger_{+1}\psi^\dagger_{-1}\psi_{0}\psi_{0}+\psi^\dagger_{0}\psi^\dagger_{0}\psi_{+1}\psi_{-1}\right)\right].
\end{eqnarray}
For two spin-1 atoms the projection operator to total spin F can be written as
\begin{eqnarray}
P_0=\frac{1-S_1\cdot S_2}{3}, P_2=\frac{2+S_1\cdot S_2}{3}.
\end{eqnarray}
Thus the interaction can be written using the spin operators
\begin{eqnarray}
V=g_0P_0+g_2P_2=c_0+c_2S_1\cdot S_2
\end{eqnarray}
where
\begin{eqnarray}
c_0=\frac{g_0+2g_2}{3}, c_2=\frac{g_2-g_0}{3}.
\end{eqnarray}
For $c_2<0$ we have a ferromagnetic spinor Bose gas, and $c_2>0$ corresponds to an anti-ferromagnetic spinor Bose gas.

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Others:

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1. Atom-Light Interaction (pdf)
2. Rotating Wave Approximation (pdf)
3. Raman scattering (pdf)
4. Exact Solution of Landau-Zener transition (pdf)
5. Hyperfind structure of Lithium (hyperfineLi.nb)